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masya89 [10]
4 years ago
10

pulse train with a frequency of 1 MHz is counted using a modulo-1024 ripple-counter built with J-K flip flops. For proper operat

ion of the counter, the maximum permissible propagation delay per flip flop stage is ______ n sec.
Physics
1 answer:
omeli [17]4 years ago
4 0

Answer:

The maximum permissible propagation delay per flip flop stage is<u> 100 </u>n sec

Explanation:

1024 ripple counter has 10 J-K flip flops(210 = 1024).  

So the total delay will be 10×x where x is the delay of each J-K flip flops.

The period of the clock pulse is 1× 10⁻⁶ s.

Now

10x <= 10⁻⁶ s

x <= 100 ns

x= 100 ns for prpoer operation.

pulse train with a frequency of 1 MHz is counted using a modulo-1024 ripple-counter built with J-K flip flops. For proper operation of the counter, the maximum permissible propagation delay per flip flop stage is <u>100 </u>n sec.

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Induced EMF and Current in a Shrinking LoopShrinking Loop. A circular loop of flexible iron wire has an initial circumference of
vodomira [7]

Answer:

Explanation:

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3 years ago
A 1.0-cm-tall object is 13 cm in front of a converging lens that has a 40 cm focal length.
kicyunya [14]

A) Image position: -19.3 cm

B) Image height: 1.5 cm, upright

Explanation:

A)

In order to calculate the image position, we can use the lens equation:

\frac{1}{p}+\frac{1}{q}=\frac{1}{f}

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

In this problem, we have:

p = 13 cm (object distance)

f = 40 cm (focal length, positive for a converging lens)

So the image distance is

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{40}-\frac{1}{13}=-0.0519\\q=\frac{1}{-0.0519}=-19.3 cm

The negative sign means that the image is virtual.

B)

In order to calculate the image height, we use the magnification equation:

\frac{y'}{y}=-\frac{q}{p}

where

y' is the image height

y is the object height

In this problem, we have:

y = 1.0 cm (object height)

p = 13 cm

q = -19.3 cm

Therefore, the image heigth is

y'=-\frac{qy}{p}=-\frac{(-19.3)(1.0)}{13}=1.5 cm

And the positive sign means the image is upright.

6 0
3 years ago
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denis-greek [22]

Answer:

<h2>4.6 m/s²</h2>

Explanation:

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v is the final velocity

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a is the acceleration

Since the body is from rest u = 0

From the question we have

a =  \frac{23 - 0}{5}  =  \frac{23}{5}  \\

We have the final answer as

<h3>4.6 m/s²</h3>

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